Maximum likelihood estimator for uniform distribution $U(-\theta, 0)$

Question

Consider $X_1,X_2,...,X_n$ i.i.d $U(-\theta,0)$.

I want to find the maximum likelihood estimator of $\theta$.

I know that $f(x,\theta)=\frac{1}{\theta}$ for $-\theta < x < 0$ and that $L_n(\theta, x)= \frac{1}{\theta^n}$.

If we were looking at $U(0,\theta)$, then the MLE of $\theta$ would be $x_{(n)}$ because $L_n(\theta, x)= \frac{1}{\theta^n}$ is decreasing from $0 < x < \theta$ and would thus be maximized at the max $x_i$, which is $x_{(n)}$.

For my case, since $L_n(\theta, x)= \frac{1}{\theta^n}$ is an increasing function for $-\theta < x < 0$, then $L_n(\theta, x)= \frac{1}{\theta^n}$ will be maximized at the max $x_i$, and thus the MLE of $\theta$ will be $x_{(n)}$ as well.

I think this is correct, but it seems very silly to me that for both cases you can just say that it will be maximized at the max $x_i$. Could someone better explain this to me?

Answer 1

Note that the likelihood function is a function of $\theta$. In particular, $$L_n(\theta;\vec X) = \left \{ \begin{matrix}\frac{1}{\theta^n} & \text{if $\theta \ge -X_i$ for $i=1,2,\cdots, n$,} \\ 0 & \text{otherwise.}\end{matrix}\right.$$ Here $\theta \ge -X_i$ comes from $-\theta \le X_i$. Now, $L_n(\theta,\vec X)$ is a decreasing function of $\theta$. Consequently, $L_n(\theta;\vec x)$ attains its maximum when $\theta = \max\{-X_i\}=-\min\{X_i\}=-X_{(1)}$.

Answer 2

Note that all $X_i$ are negative, but you may want $\theta$ to be positive. So you cannot choose $X_{(n)}$ as the MLE because that would give a negative MLE.

If $-\theta > X_{(1)} $, then the likelihood will be zero because $X_{(1)}$ is not in the support of the distribution. If $-\theta \leq X_{(1)} $, then the likelihood will be $\frac{1}{\theta^n}$. So the maximum likelihood is attained at $\theta = - X_{(1)}$.

Answer 3

Your reasoning for the $U(0,\theta)$ case is wrong, so is interfering with the $U(-\theta,0)$ case.

In the $U0,\theta)$ case the likelihood (which is a function of $\theta$ !) is $\frac{1}{\theta^n}$ if $\theta\ge x_{(n)}$ and zero otherwise since the support requires all $x_i <\theta.$ Since $1/\theta^n$ is decreasing in $\theta,$ you must pick the smallest value of $\theta$ that is $\ge x_{(n)},$ i.e. $x_{(n)}.$

Likewise for $U(-\theta,0),$ the likelihood is $1/\theta^n$ for $\theta \ge -x_{(1)}$ and zero otherwise since the support requires all $x_i \ge -\theta.$ So the MLE is $-x_{(1)}.$ This should make sense since this is just a mirror image of the previous problem.